Given the function difined by $f(x)=x^2-2x+1$
Fin real $k$ such that:
For every $x \in(2,4)$ : $|f(x)-4|\leq k|x-3|$
Fin $\delta$ such that :
$|x-3|<\delta$ => |$f(x)-4|<\epsilon$
The deduce the limit of $f(x)$ in the point $3$
$|x^2-2x+1-4|=|x^2-2x-3|=|x-3||x+1|$
$x \in (2,4) |x+1|\leq 5$ so $k=5$
And so :
$|x-3|<\epsilon$ and $|x+1|<5$ so
$\delta=\inf(\frac{\epsilon}{5};1)$
Does this have any relation with MVT (mean value theorem)